According to the orbital concept, each carbon atom in benzene is $$s{p^2}$$ - hybridised and one $$p$$ - orbital of each carbon remains unhybridised. The $$\pi $$ - electron charge in benzene is not confined to space between two carbon atoms as in ethylene, but is spread over a greater area. This is known as the delocalisation of the electron charge.

The reactivity depends upon the stability of the carbocation formed by addition of \[{{H}^{+}}.\]  The carbocations resulting from alkenes ( I, II and III ) are respectively

Since the stability decreases in the order : $$Z > Y > X,$$   therefore, reactivity decreases in the order : III > II > I.

No explanation is given for this question. Let's discuss the answer together.

Conformations of $$n$$ - butane are as under

Staggered conformation has minimum repulsion, so it is the most stable. The order of stability is staggered > gauche > eclipsed
Energy order eclipsed > gauche > staggered

Alkynes having terminal $$ - C \equiv H$$  react with Na in liquid ammonia to yield $${H_2}$$  gas of the given compounds $$C{H_3}C{H_2}C \equiv CH$$    can react with Na in liquid $$N{H_3}$$  so the correct answer is (B).
\[C{{H}_{3}}C{{H}_{2}}C\equiv CH\xrightarrow[\text{liquid}\,\,N{{H}_{3}}]{Na\,\,\text{in}}\]       \[C{{H}_{3}}C{{H}_{2}}C\equiv {{C}^{-}}N{{a}^{+}}+\frac{1}{2}{{H}_{2}}\left( g \right)\]

The reaction is Wurtz reaction and proceeds as follows :
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}-\underset{\begin{align} & |\,\,\,\, \\ & C{{H}_{2}} \\ & |\,\,\,\, \\ & C{{H}_{3}} \\ \end{align}}{\mathop{CH}}\,-Br+2Na\,\,+\]       \[Br-\underset{\begin{align} & |\,\,\,\,\, \\ & C{{H}_{2}} \\ & |\,\,\,\,\, \\ & C{{H}_{3}} \\ \end{align}}{\mathop{CH}}\,-C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\xrightarrow{\text{ethar}}\]
\[\underset{\text{4,5-Diethyloctane}}{\mathop{\overset{8\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{3}}}}\,\overset{7\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,\overset{6\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,-\underset{\begin{align} & | \\ & C{{H}_{2}} \\ & | \\ & C{{H}_{3}} \\ \end{align}}{\overset{5\,\,\,\,\,}{\mathop{CH}}}\,-\underset{\begin{align} & | \\ & C{{H}_{2}} \\ & | \\ & C{{H}_{3}} \\ \end{align}}{\overset{4\,\,\,\,\,\,}{\mathop{CH}}}\,-\overset{3\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,\overset{2\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,\overset{1\,\,\,\,\,\,\,}{\mathop{C{{H}_{3}}}}\,}}\,\]

$${\text{Only }}\left( {\text{D}} \right){\text{ can form 3}}\,{\text{ - }}\,{\text{Octyne}}$$
\[C{{H}_{3}}C{{H}_{2}}C\equiv CH\xrightarrow[-N{{H}_{3}}]{NaN{{H}_{2}}}\]       \[C{{H}_{3}}C{{H}_{2}}C\equiv {{C}^{-}}N{{a}^{+}}\]     \[\xrightarrow[\left( {{S}_{N}}2 \right)]{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br}C{{H}_{3}}C{{H}_{2}}C\]       \[\equiv CC{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}+NaBr\]

Anti addition of hydrogen atoms to the triple bond occurs when alkynes are reduced with sodium ( or lithium ) metal in ammonia, ethylamine, or alcohol at low temperatures, This reaction called, a dissolving metal reduction, produces an $$\left( E \right) - $$ or $$trans$$  - alkene. Sodium in liq. $$N{H_3}$$  is used as a source of electrons in the reduction of an alkyne to a $$trans$$  alkene.

When $$Ca{C_2}$$  reacts with water, it gives ethyne.
$$Ca{C_2} + 2{H_2}O \to CH \equiv CH + Ca{\left( {OH} \right)_2}$$